Question

A compound was analyzed and found to be 12.1% C, 71.7% Cl, and 16.2% O. What is the empirical formula for this compound?

• A. $C_{2}OCl$
• B. $COCl$
• C. $C{O}_{2}C{l}_2$
• D. $C_2{O}_{2}Cl$
• E. $CC{l}_{2}O$

Answer 1

The correct option is E $CC{l}_{2}O$

The atomic amsses of $C,Cl$ and $O$ are 12 g/mol, 35.5 g/mol and 16 g/mol respectively. 100 g of compound contains 12.1 g C, 71.7 g Cl and 16.2 g O.
The number of moles is the ratio of mass to molar mass.
The number of moles of C $=\frac{12.1g}{12g/mol}\simeq 1mol$

The number of moles of Cl $=\frac{71.7}{35.5}\simeq 2mol$

The number of moles of O $=\frac{16.2g}{16g/mol}\simeq 1$
Hence, the ratio of number of moles of C, Cl and O is $1:2:1$
The empirical formula of the compound is $CC{l}_{2}O$